Solution:

Answer yes, no

The obvious approach is to make the triangle isosceles. So suppose the sides are $a$, $b$, $b$. Then the length $m$ of a median to one of the sides length $b$ satisfies: $a^2 + b^2 = 2m^2 + b^2/2$. The simplest possibility is to take $m = b$, so $a^2 = 3b^2/2$. Thus if $b = 2$, $a = \sqrt{6}$.

Suppose we have a triangle $ABC$, with medians $AD$, $BE$, $CF$, and $BC/AD$, $CA/BE$, $AB/CF$ all integers. If $AD = BC/2$, then $\angle A = 90^\circ$. If $AD < BC/2$, then $\angle A$ is obtuse, so at least two of the medians must be equal to the corresponding sides. So wlog we have $b^2 + c^2 = 5a^2/2$, $c^2 + a^2 = 5b^2/2$. Subtracting, $b^2 - a^2 = (5/2)(a^2 - b^2)$, so $a = b$. Hence $c/a = \sqrt{3/2}$. So the third median has length $m$ where $a^2 + a^2 = (3/4)a^2 + 2m^2$, so $a/m = \sqrt{8/5}$, which is not integral. Contradiction.